STEP3 2019 -- Pure Mathematics

Exam: STEP3 | Year: 2019 | Questions: Q1—Q8 | Total marks per question: 20

All questions are pure mathematics. Solutions and examiner commentary are included below.

Overview

Q	Topic	Difficulty	Key Techniques
1	纯数	Challenging	矩阵特征值法,消元法解ODE,相平面分析,极值点定位
2	纯数	Challenging	极限定义推导,函数方程微分法,分离变量,双曲函数恒等式
3	纯数	Challenging	线性代数,不变子空间理论,分类讨论,方程组求解
4	纯数	Hard	Vieta公式,平方和非负性分析,整数约束下的穷举,反证法
5	纯数	Hard	代换积分法,部分分式分解,反三角函数,积分限变换
6	纯数	Challenging	复数模运算,反演变换几何性质,圆的方程分析,分类讨论
7	纯数	Challenging	隐式方程分析,判别式法,导数求切线方向,曲线对称性
8	纯数	Challenging	法向量计算,向量叉积,三角恒等式推导,不等式证明

Question 1

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

1 The coordinates of a particle at time $t$ are $x$ and $y$ . For $t \geqslant 0$ , they satisfy the pair of coupled differential equations

$\dot{x} = -x - ky$ $\dot{y} = x - y$

where $k$ is a constant. When $t = 0$ , $x = 1$ and $y = 0$ .

(i) Let $k = 1$ . Find $x$ and $y$ in terms of $t$ and sketch $y$ as a function of $t$ .

Sketch the path of the particle in the $x$ - $y$ plane, giving the coordinates of the point at which $y$ is greatest and the coordinates of the point at which $x$ is least.

(ii) Instead, let $k = 0$ . Find $x$ and $y$ in terms of $t$ and sketch the path of the particle in the $x$ - $y$ plane.

Hint

(i) $\dot{x} = -x - y$

So $y = -\dot{x} - x$

As $\dot{y} = x - y$ , $-\ddot{x} - \dot{x} = x + \dot{x} + x$

$\ddot{x} + 2\dot{x} + 2x = 0$

[M1]

AQE $\lambda^2 + 2\lambda + 2 = 0$ , $\lambda = -1 \pm i$ so $x = e^{-t}(A \cos t + B \sin t)$ [M1 A1 cao]

[Alternatively, $x = \dot{y} + y$ leading to $\ddot{y} + 2\dot{y} + 2y = 0$ and $y = e^{-t}(C \cos t + D \sin t)$ etc]

So $y = e^{-t}(A \cos t + B \sin t) - e^{-t}(-A \sin t + B \cos t) - e^{-t}(A \cos t + B \sin t)$

$= e^{-t}(A \sin t - B \cos t)$

$t = 0, x = 1 \Rightarrow A = 1$ and $t = 0, y = 0 \Rightarrow B = 0$

So $x = e^{-t} \cos t$ and $y = e^{-t} \sin t$ [M1 A1 cao]

$t$	$y$
$0$	$0$
$\pi$	$0$
$2\pi$	$0$

G1 G1 [7]

$y$ is greatest when $\dot{y} = 0 \Rightarrow x = y \Rightarrow \tan t = 1 \Rightarrow t = \frac{\pi}{4} + n\pi$

Hence $t = \frac{\pi}{4}$ thus $\left( \frac{e^{-\frac{\pi}{4}}}{\sqrt{2}}, \frac{e^{-\frac{\pi}{4}}}{\sqrt{2}} \right)$ [M1]

$x$ is least when $\dot{x} = 0 \Rightarrow x = -y \Rightarrow \tan t = -1 \Rightarrow t = \frac{3\pi}{4} + n\pi$

Hence $t = \frac{3\pi}{4}$ thus $\left( -\frac{e^{-\frac{3\pi}{4}}}{\sqrt{2}}, \frac{e^{-\frac{3\pi}{4}}}{\sqrt{2}} \right)$ [M1]

G1 G1 G1 [5]

(ii) $\dot{x} = -x \implies x = Ae^{-t}$

$t = 0, x = 1 \implies A = 1 \implies x = e^{-t}$ M1

So $\dot{y} + y = e^{-t}$

The integrating factor is $e^t$ . Thus $e^t y = \int 1 dt = t + c$

$y = (t + c)e^{-t}$ M1

$t = 0, y = 0 \implies c = 0$ so $y = te^{-t}$ A1 cao

$\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{(1-t)e^{-t}}{-e^{-t}} = t - 1$ so there is a maximum at $(e^{-1}, e^{-1})$ M1

As $t \to \infty$ , $\frac{dy}{dx} \to \infty$ M1

Alternatively,

As $\dot{y} = x - y$ , $\ddot{y} = \dot{x} - \dot{y} = -x - \dot{y}$

So $\ddot{y} + 2\dot{y} + y = 0$ yielding $y = (At + B)e^{-t}$ M1

$x$	$y$
$0$	$0$
$e^{-1}$	$e^{-1}$
$1$	$0$

G1 G1 G1 [8]

Model Solution

Part (i) ( $k = 1$ )

With $k = 1$ , the system becomes $\dot{x} = -x - y$ and $\dot{y} = x - y$ .

From the first equation, $y = -\dot{x} - x$ , so $\dot{y} = -\ddot{x} - \dot{x}$ .

Substituting into the second equation:

$-\ddot{x} - \dot{x} = x - (-\dot{x} - x) = 2x + \dot{x}$

$\ddot{x} + 2\dot{x} + 2x = 0$

The auxiliary equation is $\lambda^2 + 2\lambda + 2 = 0$ , giving $\lambda = \dfrac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$ .

So $x = e^{-t}(A\cos t + B\sin t)$ .

To find $y$ : since $y = -\dot{x} - x$ , we compute $\dot{x}$ :

$\dot{x} = -e^{-t}(A\cos t + B\sin t) + e^{-t}(-A\sin t + B\cos t)$

$= e^{-t}[(-A+B)\cos t + (-A-B)\sin t]$

Therefore:

$y = -\dot{x} - x = -e^{-t}[(-A+B)\cos t + (-A-B)\sin t] - e^{-t}(A\cos t + B\sin t)$

$= e^{-t}[(A - B)\cos t + (A + B)\sin t - A\cos t - B\sin t]$

$= e^{-t}(-B\cos t + A\sin t)$

Applying initial conditions: $x(0) = 1$ gives $A = 1$ , and $y(0) = 0$ gives $-B = 0$ , so $B = 0$ .

$\boxed{x = e^{-t}\cos t, \qquad y = e^{-t}\sin t}$

Sketch of $y$ as a function of $t$ :

$y(0) = 0$ , and $y = 0$ whenever $\sin t = 0$ , i.e. at $t = 0, \pi, 2\pi, \ldots$ The envelope $\pm e^{-t}$ decays exponentially. $y$ reaches its first positive maximum at $t = \frac{\pi}{4}$ , its first negative minimum at $t = \frac{5\pi}{4}$ , and the oscillations die away as $t \to \infty$ .

Path in the $x$ - $y$ plane:

Note that $x^2 + y^2 = e^{-2t}(\cos^2 t + \sin^2 t) = e^{-2t}$ , so the distance from the origin is $r = e^{-t}$ , which decreases monotonically. The particle spirals inward toward the origin.

Maximum $y$ : We need $\dot{y} = 0$ , i.e. $x - y = 0$ , so $x = y$ :

$e^{-t}\cos t = e^{-t}\sin t \implies \tan t = 1 \implies t = \frac{\pi}{4} + n\pi$

For the first (and largest) positive $y$ , $t = \frac{\pi}{4}$ :

$\left(\frac{e^{-\pi/4}}{\sqrt{2}},\; \frac{e^{-\pi/4}}{\sqrt{2}}\right)$

Minimum $x$ : We need $\dot{x} = 0$ , i.e. $-x - y = 0$ , so $x = -y$ :

$e^{-t}\cos t = -e^{-t}\sin t \implies \tan t = -1 \implies t = \frac{3\pi}{4} + n\pi$

For the first such value, $t = \frac{3\pi}{4}$ :

$\left(-\frac{e^{-3\pi/4}}{\sqrt{2}},\; \frac{e^{-3\pi/4}}{\sqrt{2}}\right)$

Part (ii) ( $k = 0$ )

With $k = 0$ , $\dot{x} = -x$ and $\dot{y} = x - y$ .

From the first equation: $\dot{x} = -x$ gives $x = Ae^{-t}$ . With $x(0) = 1$ : $A = 1$ , so $x = e^{-t}$ .

Substituting into the second equation: $\dot{y} + y = e^{-t}$ .

The integrating factor is $e^t$ :

$\frac{d}{dt}(e^t y) = e^t \cdot e^{-t} = 1$

$e^t y = t + c$

With $y(0) = 0$ : $c = 0$ , so:

$\boxed{x = e^{-t}, \qquad y = te^{-t}}$

Path in the $x$ - $y$ plane:

The path starts at $(1, 0)$ when $t = 0$ . As $t$ increases, $x = e^{-t}$ decreases monotonically while $y = te^{-t}$ increases from 0 to a maximum and then decreases back to 0.

To find the maximum $y$ on the curve, we compute $\dfrac{dy}{dx} = \dfrac{\dot{y}}{\dot{x}} = \dfrac{(1-t)e^{-t}}{-e^{-t}} = t - 1$ .

Setting $\dfrac{dy}{dx} = 0$ : $t = 1$ , giving the point $(e^{-1}, e^{-1})$ .

As $t \to \infty$ , $\dfrac{dy}{dx} = t - 1 \to \infty$ , meaning the tangent becomes vertical (approaching the origin along the positive $x$ -axis from above).

The path starts at $(1,0)$ , rises to $(e^{-1}, e^{-1})$ , then curves back toward the origin.

Examiner Notes

k=0时方程退化为可分离变量形式,需注意;画轨迹图时需准确标注y最大值和x最小值的坐标。耦合ODE是STEP3高频考点。

Question 2

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

2 The definition of the derivative $f'$ of a (differentiable) function $f$ is

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} . \qquad \text{(*)}$

(i) The function $f$ has derivative $f'$ and satisfies

$f(x + y) = f(x)f(y)$

for all $x$ and $y$ , and $f'(0) = k$ where $k \neq 0$ . Show that $f(0) = 1$ .

Using ( $*$ ), show that $f'(x) = kf(x)$ and find $f(x)$ in terms of $x$ and $k$ .

(ii) The function $g$ has derivative $g'$ and satisfies

$g(x + y) = \frac{g(x) + g(y)}{1 + g(x)g(y)}$

for all $x$ and $y$ , $|g(x)| < 1$ for all $x$ , and $g'(0) = k$ where $k \neq 0$ .

Find $g'(x)$ in terms of $g(x)$ and $k$ , and hence find $g(x)$ in terms of $x$ and $k$ .

Hint

(i) Let $y = 0$ , $f(x + 0) = f(x)f(0) \forall x$ , so $f(x) = f(x)f(0) \forall x$ M1

Thus $f(x)(1 - f(0)) = 0 \forall x$ . Therefore, $f(x) = 0 \forall x$ , or $f(0) = 1$ M1

If $f(x) = 0 \forall x$ then $f'(x) = 0 \forall x$ but $f'(0) = k \neq 0$ so $f(0) = 1$ A1* cso

$f'(x) = \lim_{h \to 0} \left( \frac{f(x+h) - f(x)}{h} \right) = \lim_{h \to 0} \left( \frac{f(x)f(h) - f(x)}{h} \right) = f(x) \lim_{h \to 0} \left( \frac{f(h) - 1}{h} \right)$ M1

But $\lim_{h \to 0} \left( \frac{f(h) - 1}{h} \right) = \lim_{h \to 0} \left( \frac{f(0+h) - f(0)}{h} \right) = f'(0) = k$ so $f'(x) = kf(x)$ M1 A1* cso

Thus

$\frac{f'(x)}{f(x)} = k$

Integrating

$\ln(f(x)) = kx + c$

So $f(x) = e^{kx+c} = Ae^{kx}$ . As $f(0) = 1$ , $A = 1$ , so $f(x) = e^{kx}$ M1 A1 cao [9]

(ii) Let $y = 0$ ,

$g(x + 0) = \frac{g(x) + g(0)}{1 + g(x)g(0)} \forall x$

Thus $g(x) + (g(x))^2 g(0) = g(x) + g(0) \forall x$

So $((g(x))^2 - 1) g(0) = 0 \forall x$ M1

As $|g(x)| < 1$ , $(g(x))^2 - 1 \neq 0$ , and thus $g(0) = 0$ A1

$g'(x) = \lim_{h \to 0} \left( \frac{g(x + h) - g(x)}{h} \right)$

$= \lim_{h \to 0} \left( \frac{\frac{g(x) + g(h)}{1 + g(x)g(h)} - g(x)}{h} \right) = \lim_{h \to 0} \left( \frac{g(x) + g(h) - g(x)(1 + g(x)g(h))}{h(1 + g(x)g(h))} \right)$

$= \lim_{h \to 0} \left( \frac{g(h) (1 - (g(x))^2)}{h(1 + g(x)g(h))} \right) = (1 - (g(x))^2) \lim_{h \to 0} \left( \frac{g(h)}{h(1 + g(x)g(h))} \right)$

$\lim_{h \to 0} \left( \frac{g(h)}{h(1 + g(x)g(h))} \right) = \lim_{h \to 0} \left( \frac{g(h)/h}{(1 + g(x)g(h))} \right) = \lim_{h \to 0} \left( \frac{g(h)}{h} \right) = g'(0) = k$

M1 A1

Thus $g'(x) = k(1 - (g(x))^2)$ A1

$\frac{g'(x)}{(1 - (g(x))^2)} = k$

Integrating,

$\tanh^{-1}(g(x)) = kx + c$

$g(x) = \tanh(kx + c)$

As $g(0) = 0$ , $\tanh(c) = 0$ and so $c = 0$ M1

Thus

$g(x) = \tanh(kx)$

A1 [11]

[Alternatively, integrating having used partial fractions,

$\frac{1}{2} \ln \left( \frac{1 + g(x)}{1 - g(x)} \right) = kx + c$ M1

As $g(0) = 0$ , $c = 0$

$\frac{1 + g(x)}{1 - g(x)} = e^{2kx}$ M1

and so

$g(x) = \frac{e^{2kx} - 1}{e^{2kx} + 1}$ A1 [3]

Model Solution

Part (i)

Showing $f(0) = 1$ :

Set $y = 0$ in $f(x+y) = f(x)f(y)$ :

$f(x) = f(x)f(0) \quad \text{for all } x$

$f(x)(1 - f(0)) = 0 \quad \text{for all } x$

So either $f(x) = 0$ for all $x$ , or $f(0) = 1$ .

If $f(x) = 0$ for all $x$ , then $f'(x) = 0$ for all $x$ , contradicting $f'(0) = k \neq 0$ .

Therefore $f(0) = 1$ .

Showing $f'(x) = kf(x)$ :

Using the definition $(*)$ :

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Applying the functional equation with $f(x+h) = f(x)f(h)$ :

$f'(x) = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$

Now, since $f(0) = 1$ :

$\lim_{h \to 0} \frac{f(h) - 1}{h} = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} = f'(0) = k$

Therefore $f'(x) = kf(x)$ .

Finding $f(x)$ :

$\frac{f'(x)}{f(x)} = k$

Integrating both sides:

$\ln f(x) = kx + c$

$f(x) = Ae^{kx}$

Since $f(0) = 1$ : $A = 1$ , so:

$\boxed{f(x) = e^{kx}}$

Part (ii)

Finding $g(0)$ :

Set $y = 0$ in $g(x+y) = \dfrac{g(x)+g(y)}{1+g(x)g(y)}$ :

$g(x) = \frac{g(x) + g(0)}{1 + g(x)g(0)}$

$g(x)(1 + g(x)g(0)) = g(x) + g(0)$

$g(x) + g(x)^2 g(0) = g(x) + g(0)$

$g(0)(g(x)^2 - 1) = 0$

Since $|g(x)| < 1$ for all $x$ , we have $g(x)^2 - 1 \neq 0$ , so $g(0) = 0$ .

Finding $g'(x)$ :

$g'(x) = \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} = \lim_{h \to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)} - g(x)}{h}$

Combining the fraction in the numerator:

$= \lim_{h \to 0} \frac{g(x) + g(h) - g(x)(1 + g(x)g(h))}{h(1 + g(x)g(h))}$

$= \lim_{h \to 0} \frac{g(h) - g(x)^2 g(h)}{h(1 + g(x)g(h))}$

$= \lim_{h \to 0} \frac{g(h)(1 - g(x)^2)}{h(1 + g(x)g(h))}$

$= (1 - g(x)^2) \lim_{h \to 0} \frac{g(h)/h}{1 + g(x)g(h)}$

As $h \to 0$ : $g(h) \to g(0) = 0$ , so $1 + g(x)g(h) \to 1$ , and $\dfrac{g(h)}{h} \to g'(0) = k$ .

$\boxed{g'(x) = k(1 - g(x)^2)}$

Finding $g(x)$ :

$\frac{g'(x)}{1 - g(x)^2} = k$

Integrating both sides. Since $|g(x)| < 1$ , we use the standard result $\displaystyle\int \frac{du}{1-u^2} = \tanh^{-1} u$ :

$\tanh^{-1}(g(x)) = kx + c$

With $g(0) = 0$ : $\tanh^{-1}(0) = 0$ , so $c = 0$ .

$\boxed{g(x) = \tanh(kx)}$

We can verify: $\tanh$ satisfies the addition formula $\tanh(x+y) = \dfrac{\tanh x + \tanh y}{1 + \tanh x \tanh y}$ , $|\tanh(kx)| < 1$ for all $x$ , and $\dfrac{d}{dx}\tanh(kx)\big|_{x=0} = k\,\text{sech}^2(0) = k$ . All conditions are satisfied.

Examiner Notes

第(i)部分需先证明f(0)=1;第(ii)部分g的函数方程形式类似tanh加法公式,推导g’(x)后需正确积分得到g(x)=tanh(kx)。

Question 3

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

3 The matrix A is given by $\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.$

(i) You are given that the transformation represented by A has a line $L_1$ of invariant points (so that each point on $L_1$ is transformed to itself). Let $(x, y)$ be a point on $L_1$ . Show that $((a - 1)(d - 1) - bc)xy = 0$ .

Show further that $(a - 1)(d - 1) = bc$ .

What can be said about A if $L_1$ does not pass through the origin?

(ii) By considering the cases $b \neq 0$ and $b = 0$ separately, show that if $(a - 1)(d - 1) = bc$ then the transformation represented by A has a line of invariant points. You should identify the line in the different cases that arise.

(iii) You are given instead that the transformation represented by A has an invariant line $L_2$ (so that each point on $L_2$ is transformed to a point on $L_2$ ) and that $L_2$ does not pass through the origin. If $L_2$ has the form $y = mx + k$ , show that $(a - 1)(d - 1) = bc$ .

Hint

(i) $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \Rightarrow ax + by = x \quad cx + dy = y$ M1

$(a - 1)x = -by$

$-cx = (d - 1)y$

Thus $(a - 1)x(d - 1)y = bycx$ , that is $(a - 1)(d - 1)xy - bcxy = 0$

So $((a - 1)(d - 1) - bc)xy = 0$ M1

Thus $((a - 1)(d - 1) - bc) = 0$ or $x = 0$ or $y = 0$ M1

If $L_1$ is $x = 0$ , then both $by = 0$ and $dy = y, \forall y$

Thus, $b = 0$ and $d = 1$ , meaning that $((a - 1)(d - 1) - bc) = 0$ E1

Similarly, if $L_1$ is $y = 0$ , then both $cx = 0$ and $ax = x, \forall x$

Then $c = 0$ and $a = 1$ , meaning that $((a - 1)(d - 1) - bc) = 0$

In all three cases, $(a - 1)(d - 1) = bc$ E1

Alternatively,

$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$ M1

So $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

That is $\begin{pmatrix} a - 1 & b \\ c & d - 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$ M1

As this is true for a line of invariant points, it does not have a unique solution and so E1

$det \begin{pmatrix} a - 1 & b \\ c & d - 1 \end{pmatrix} = 0$ M1

and so $((a - 1)(d - 1) - bc) = 0$ which implies both $((a - 1)(d - 1) - bc)xy = 0$ and $(a - 1)(d - 1) = bc$ E1

If $L_1$ does not pass through the origin then either $L_1$ is a) $y = mx + k$ with $k \neq 0$

or b) $x = k$ with $k \neq 0$ E1

For a) $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} = \begin{pmatrix} x \\ mx + k \end{pmatrix} \quad \forall x$

Thus $ax + b(mx + k) = x$ and $cx + d(mx + k) = mx + k$

As these apply for all $x$ , and $k \neq 0$ , $bk = 0$ which implies $b = 0$ and $a + bm = 1$ and thus $a = 1$

Also $dk = k$ implying $d = 1$ and $c + dm = m$ which thus gives $c = 0$ M1

For b) $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} k \\ y \end{pmatrix} = \begin{pmatrix} k \\ y \end{pmatrix} \forall y$

Thus $ak + by = k$ and $ck + dy = y$

So $ak = k$ implying $a = 1$ and $b = 0$ and $ck = 0$ implying $c = 0$ and $d = 1$ M1

Thus $\mathbf{A} = \mathbf{I}$ A1 [9]

(ii) If $(a - 1)(d - 1) = bc$ and $b \neq 0$

then $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}$ is an invariant point iff $ax + by = x$ and $cx + dy = y$ M1

That is $(a - 1)x + by = 0$ , $(a - 1)(d - 1)x + b(d - 1)y = 0$ , $bcx + b(d - 1)y = 0$ and so $cx + (d - 1)y = 0$ as required. E1

The line of invariant points is thus $(a - 1)x + by = 0$ which is $cx + (d - 1)y = 0$ A1

If $(a - 1)(d - 1) = bc$ and $b = 0$ then $a = 1$ or if $a \neq 1$ , $d = 1$

$a = 1$ , $\begin{pmatrix} 1 & 0 \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ cx + dy \end{pmatrix}$ so points on $cx + (d - 1)y = 0$ are invariant. B1

$a \neq 1$ , $\begin{pmatrix} a & 0 \\ c & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax \\ cx + y \end{pmatrix}$ so points on $x = 0$ are invariant. B1 [5]

(iii) $L_2$ is an invariant line implies $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ mx + k \end{pmatrix} = \begin{pmatrix} x' \\ mx' + k \end{pmatrix}$ and as $L_2$ does not pass through the origin $k \neq 0$ M1

So $ax + b(mx + k) = x'$ and $cx + d(mx + k) = mx' + k$

As these are true for all $x$ , true for $x = 0$ and thus $bk = x'$ and $dk = mx' + k$ giving

$dk = mbk + k$ and as $k \neq 0$ , $d = mb + 1$ E1

Similarly, for $x = 1$ and thus $a + b(m + k) = x''$ and $c + d(m + k) = mx'' + k$

So $c + d(m + k) = m(a + b(m + k)) + k$

$c + dm + dk = ma + (m + k)(d - 1) + k$

$c + dm + dk = ma + dm + dk - m - k + k$

$c = ma - m$ E1

So $m(a - 1) = c$ and $d - 1 = mb$

Hence, multiplying these $m(a - 1)(d - 1) = mbc$ E1

Thus, if $m \neq 0$ , $(a - 1)(d - 1) = bc$

If $m = 0$ , $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ k \end{pmatrix} = \begin{pmatrix} x' \\ k \end{pmatrix}$ giving $ax + bk = x'$ and $cx + dk = k$ E1

As these must be true for all $x$ , when $x = 0$ , $dk = k$ giving $d = 1$ and so for choosing any $x \neq 0$ , we find $c = 0$ . Thus $(a - 1)(d - 1) = 0$ and $bc = 0$ giving $(a - 1)(d - 1) = bc$

E1 [6]

Model Solution

Part (i)

Let $(x, y)$ be a point on $L_1$ . Since it is an invariant point, $\mathbf{A}$ maps $(x, y)$ to itself:

$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}$

This gives the two equations:

$ax + by = x \qquad \Rightarrow \qquad (a-1)x = -by \qquad \qquad \text{(1)}$

$cx + dy = y \qquad \Rightarrow \qquad cx = -(d-1)y \qquad \qquad \text{(2)}$

From (1): $(a-1)x = -by$ . From (2): $(d-1)y = -cx$ .

Multiplying these two equations:

$(a-1)(d-1)xy = (-by)(-cx) = bcxy$

$((a-1)(d-1) - bc)xy = 0$

$\boxed{((a-1)(d-1) - bc)xy = 0}$

as required. $\qquad \blacksquare$

Showing $(a-1)(d-1) = bc$ :

The equation $((a-1)(d-1) - bc)xy = 0$ holds for every point on $L_1$ . Since $L_1$ is a line (not just the origin), it contains infinitely many points. We consider three cases:

Case 1: If there exists a point on $L_1$ with $x \neq 0$ and $y \neq 0$ , then we can divide by $xy$ to get $(a-1)(d-1) - bc = 0$ , i.e.\ $(a-1)(d-1) = bc$ .

Case 2: $L_1$ is the line $x = 0$ (the $y$ -axis). Then every point $(0, y)$ is invariant. From (1): $by = 0$ for all $y$ , giving $b = 0$ . From (2): $dy = y$ for all $y$ , giving $d = 1$ . Hence $(a-1)(d-1) = (a-1)(0) = 0$ and $bc = 0$ , so $(a-1)(d-1) = bc$ .

Case 3: $L_1$ is the line $y = 0$ (the $x$ -axis). Then every point $(x, 0)$ is invariant. From (1): $(a-1)x = 0$ for all $x$ , giving $a = 1$ . From (2): $cx = 0$ for all $x$ , giving $c = 0$ . Hence $(a-1)(d-1) = 0$ and $bc = 0$ , so again $(a-1)(d-1) = bc$ .

In all cases, $(a-1)(d-1) = bc$ . $\qquad \blacksquare$

If $L_1$ does not pass through the origin: Suppose $L_1$ is the line $y = mx + k$ with $k \neq 0$ . Every point $(x, mx+k)$ satisfies $\mathbf{A}\binom{x}{mx+k} = \binom{x}{mx+k}$ , so:

$(a-1)x + b(mx + k) = 0 \qquad \text{for all } x$

The constant term gives $bk = 0$ . Since $k \neq 0$ , we get $b = 0$ . Then the $x$ -coefficient gives $a - 1 + bm = 0$ , so $a = 1$ .

$(d-1)(mx + k) + cx = 0 \qquad \text{for all } x$

The constant term gives $(d-1)k = 0$ , so $d = 1$ . The $x$ -coefficient gives $c + (d-1)m = 0$ , so $c = 0$ .

Therefore $\mathbf{A} = \mathbf{I}$ . The only matrix with an invariant line not through the origin is the identity matrix.

Part (ii)

Case $b \neq 0$ : We are given $(a-1)(d-1) = bc$ . Consider the line $(a-1)x + by = 0$ , i.e.\ $y = -\frac{(a-1)}{b}x$ .

For any point $(x, y)$ on this line:

First component: $ax + by = ax + b \cdot \left(-\frac{a-1}{b}x\right) = ax - (a-1)x = x$ . ✓
Second component: $cx + dy = cx - \frac{d(a-1)}{b}x$ . We need this to equal $y = -\frac{a-1}{b}x$ , so:

$cx - \frac{d(a-1)}{b}x = -\frac{a-1}{b}x$

Dividing by $x$ (for $x \neq 0$ , and the case $x = 0$ gives $y = 0$ which is trivially invariant):

$c - \frac{d(a-1)}{b} = -\frac{a-1}{b}$

$cb - d(a-1) = -(a-1)$

$bc = (a-1)(d-1)$

This is precisely our hypothesis. So the line of invariant points is $(a-1)x + by = 0$ , equivalently $cx + (d-1)y = 0$ (these represent the same line since $(a-1)(d-1) = bc$ ).

Case $b = 0$ : The condition $(a-1)(d-1) = bc = 0$ means $a = 1$ or $d = 1$ .

Subcase $a = 1$ : $\mathbf{A} = \begin{pmatrix} 1 & 0 \\ c & d \end{pmatrix}$ . The invariant point equations become $0 = 0$ and $cx + (d-1)y = 0$ . The line of invariant points is $cx + (d-1)y = 0$ .
Subcase $d = 1$ , $a \neq 1$ : $\mathbf{A} = \begin{pmatrix} a & 0 \\ c & 1 \end{pmatrix}$ . The invariant point equations become $(a-1)x = 0$ and $cx = 0$ . Since $a \neq 1$ , we need $x = 0$ (and $cx = 0$ is then automatic). The line of invariant points is $x = 0$ .

In every case, $\mathbf{A}$ has a line of invariant points. $\qquad \blacksquare$

Part (iii)

$L_2: y = mx + k$ with $k \neq 0$ is an invariant line, meaning $\mathbf{A}$ maps every point on $L_2$ to a (possibly different) point on $L_2$ . For any $x$ , the image of $(x, mx+k)$ must be of the form $(x', mx'+k)$ :

$ax + b(mx + k) = x' \qquad \qquad \text{(3)}$

$cx + d(mx + k) = mx' + k \qquad \qquad \text{(4)}$

Case $m \neq 0$ : From (3): $x' = (a + bm)x + bk$ . Substitute into (4):

$cx + dmx + dk = m(a + bm)x + mbk + k$

$(c + dm)x + dk = (ma + m^2b)x + mbk + k$

Equating constant terms ( $x = 0$ ):

$dk = mbk + k$

Since $k \neq 0$ , divide by $k$ :

$d = mb + 1 \qquad \Rightarrow \qquad d - 1 = mb \qquad \qquad \text{(5)}$

Equating coefficients of $x$ :

$c + dm = ma + m^2b$

Using (5), $dm = mb^2 + m$ :

$c + mb^2 + m = ma + m^2b$

$c = ma - m = m(a - 1) \qquad \qquad \text{(6)}$

Multiplying (5) and (6):

$(d-1) \cdot m(a-1) = mb \cdot c$

$m(a-1)(d-1) = mbc$

Since $m \neq 0$ , divide by $m$ :

$\boxed{(a-1)(d-1) = bc} \qquad \blacksquare$

Case $m = 0$ : The line is $y = k$ with $k \neq 0$ . For any $(x, k)$ , the image is $(x', k)$ :

$ax + bk = x' \qquad \qquad \text{(7)}$

$cx + dk = k \qquad \qquad \text{(8)}$

From (8), since this must hold for all $x$ : the coefficient of $x$ gives $c = 0$ , and the constant term gives $dk = k$ , so $d = 1$ (since $k \neq 0$ ).

Therefore $(a-1)(d-1) = (a-1) \cdot 0 = 0$ and $bc = b \cdot 0 = 0$ , giving $(a-1)(d-1) = bc$ . $\qquad \blacksquare$

Examiner Notes

需区分”不变点”(fixed point)和”不变线”(invariant line)两个概念;b=0时需仔细讨论矩阵的特殊形式;第(iii)部分不变线不过原点的条件是关键。

Question 4

Topic: 纯数 | Difficulty: Hard | Marks: 20

Problem

4 The $n$ th degree polynomial $\text{P}(x)$ is said to be reflexive if:

(a) $\text{P}(x)$ is of the form $x^n - a_1x^{n-1} + a_2x^{n-2} - \dots + (-1)^n a_n$ where $n \geqslant 1$ ; (b) $a_1, a_2, \dots, a_n$ are real; (c) the $n$ (not necessarily distinct) roots of the equation $\text{P}(x) = 0$ are $a_1, a_2, \dots, a_n$ .

(i) Find all reflexive polynomials of degree less than or equal to 3.

(ii) For a reflexive polynomial with $n > 3$ , show that $2a_2 = -a_2^2 - a_3^2 - \dots - a_n^2.$

Deduce that, if all the coefficients of a reflexive polynomial of degree $n$ are integers and $a_n \neq 0$ , then $n \leqslant 3$ .

(iii) Determine all reflexive polynomials with integer coefficients.

Hint

(i) Degree 1, $x - a_1 = 0$ has root $x = a_1$ and so $x - a_1$ is reflexive. [B1]

Degree 2, $x^2 - a_1x + a_2 = 0$ has to have roots $a_1$ and $a_2$ to be reflexive.

Thus ${a_1}^2 - a_1a_1 + a_2 = 0$ (giving $a_2 = 0$ ) and ${a_2}^2 - a_1a_2 + a_2 = 0$ [M1] which is consistent. Thus, $x^2 - a_1x$ (+ 0) [B1]

$a_1 = a_1 + a_2$ and $a_2 = a_1a_2$ giving $a_2 = 0$ and consistent for any $a_1$ [M1]

Thus, $x^2 - a_1x$ (+ 0) [B1]

Degree 3, $x^3 - a_1x^2 + a_2x - a_3 = 0$

$a_1 = a_1 + a_2 + a_3$

$a_2 = a_1a_2 + a_2a_3 + a_3a_1$

$a_3 = a_1a_2a_3$ [M1]

The first equation implies that $a_2 + a_3 = 0$ [M1] or in other words $a_2 = -a_3$

This result substituted into the second equation implies that $a_2 = a_2a_3$ [M1]

Continuing $a_2a_3 - a_2 = 0$ , $a_2(a_3 - 1) = 0$ [M1] so either $a_2 = 0$ and thus $a_3 = 0$ in which case the equations are consistent for any $a_1$ [M1] or $a_3 = 1$ and thus $a_2 = -1$ and $a_1 = -1$ from the third equation. [M1]

Alternatively, from the third equation, $a_3 = a_1a_2a_3$ , $a_1a_2a_3 - a_3 = 0$ , $a_3(a_1a_2 - 1) = 0$

so $a_3 = 0$ , $a_2 = 0$ and consistent for any $a_1$ or $a_1a_2 = 1$ In the latter case it is simpler to use equation two route again.

Yielding $x^3 - a_1x^2$ [A1] or $x^3 + x^2 - x - 1$ [A1]

Alternative approach ${a_1}^3 - a_1{a_1}^2 + a_2a_1 - a_3 = 0$ (A), ${a_2}^3 - a_1{a_2}^2 + a_2a_2 - a_3 = 0$ (B), and

${a_3}^3 - a_1{a_3}^2 + a_2a_3 - a_3 = 0$ (C) [M1]

So (A) implies $a_3 = a_1a_2$ and thus (B) becomes ${a_2}^3 - a_1{a_2}^2 + {a_2}^2 - a_1a_2 = 0$

Hence $a_2({a_2}^2 - a_1a_2 + a_2 - a_1) = a_2(a_2 + 1)(a_2 - a_1) = 0$ [M1]

Therefore, $a_2 = 0$ and so from (A) $a_3 = 0$ giving the polynomial $x^3 - a_1x^2$ , which is reflexive [A1]

or $a_2 = -1$ and so from (A) $a_3 = -a_1$ giving the polynomial

$x^3 - a_1x^2 - x + a_1 = (x - a_1)(x^2 - 1)$ for which the equation has roots $a_1$ and $\pm 1$ . Thus $a_3 = -a_1 = 1$ so the polynomial is $x^3 + x^2 - x - 1$ for which the equation has roots $-1$ , $-1$ and $1$ and so is reflexive M1 A1

or $a_2 = a_1$ and so A gives $a_3 = {a_1}^2$ But C is ${a_3}^3 - a_1{a_3}^2 + a_2a_3 - a_3 = 0$ and so

$a_3({a_3}^2 - a_1a_3 + a_1 - a_3) = 0$ , $a_3(a_3 - a_1)(a_3 - 1) = 0$ M1

$a_3 = 0$ , $a_1 = 0$ , $a_2 = 0$ giving the reflexive polynomial $x^3$

$a_3 = a_1$ , implies $a_1 = 0$ or $a_1 = 1$ giving $x^3$ again or $a_1 = a_2 = a_3 = 1$ giving the polynomial $x^3 - x^2 + x - 1 = (x^2 + 1)(x - 1)$ which is not reflexive. E1

$a_3 = 1$ , implies $a_1 = a_2 = \pm 1$ giving 1,1,1 which is not possible or -1,-1,1 (both already considered) E1 [11]

Summary degree 1 $x - a_1$ , degree 2 $x^2 - a_1x$ (+ 0), and degree 3 $x^3 - a_1x^2$ or $x^3 + x^2 - x - 1$

(ii)

$a_1 = \sum_{r=1}^{n} a_r$

and so,

$\sum_{r=2}^{n} a_r = 0$

$a_2 = \frac{1}{2} \sum_{i \neq j} a_i a_j$

$\left( \sum_{r=2}^{n} a_r \right)^2 = \sum_{r=2}^{n} {a_r}^2 + \sum_{i \neq j} a_i a_j - 2a_1 \sum_{r=2}^{n} a_r$

Thus

$0 = \sum_{r=2}^{n} {a_r}^2 + 2a_2$

Hence

$2a_2 = -{a_2}^2 - {a_3}^2 - \dots - {a_n}^2$

as required. A1*

${a_2}^2 + 2a_2 + 1 = 1 - {a_3}^2 - \dots - {a_n}^2$

Thus,

$1 - {a_3}^2 - \dots - {a_n}^2 \ge 0$

${a_3}^2 + \dots + {a_n}^2 \le 1$

If all the coefficients are integers, then as $a_n \neq 0$ , ${a_n}^2 = 1$ and the other coefficients for

$r = 3, \dots, n - 1$ are zero. But

$a_n = \prod_{r=1}^n a_r$

so we have established a contradiction. Thus if $a_n \neq 0$ , $n \le 3$ E1 [5]

(iii) So apart from those found in (i) with $a_1$ integer, any other reflexive polynomials must have $a_n = 0$ E1

$x^n - a_1 x^{n-1} + a_2 x^{n-2} - \dots + (-1)^{n-1} a_{n-1} x = 0$ $\therefore x(x^{n-1} - a_1 x^{n-2} + a_2 x^{n-3} - \dots + (-1)^{n-1} a_{n-1}) = 0$ M1

which gives a root of zero plus the roots of the bracketed expression. Thus we require the bracketed expression to be itself a reflexive polynomial. This can only happen if either the bracketed expression is of degree 3 or, in turn, $a_{n-1} = 0$ , and so on. E1

Hence, we have $x - a_1$ , $x^2 - a_1 x$ (+ 0), $x^3 - a_1 x^2$ , (with $a_1$ integer), $x^3 + x^2 - x - 1$ , or these multiplied by $x^r$ .

That is $(x - a_1)x^r$ or $(x + 1)^2(x - 1)x^r$ with $r = 0, 1, 2, \dots$ A1 [4]

Model Solution

Part (i)

Degree 1: $\mathrm{P}(x) = x - a_1$ . The equation $x - a_1 = 0$ has root $x = a_1$ . The single coefficient is $a_1$ , which equals the single root. So $x - a_1$ is reflexive for every real $a_1$ .

Degree 2: $\mathrm{P}(x) = x^2 - a_1 x + a_2$ . By Vieta’s formulas, the roots $a_1, a_2$ satisfy:

$a_1 + a_2 = a_1 \qquad \Rightarrow \qquad a_2 = 0 \qquad \qquad \text{(1)}$

$a_1 \cdot a_2 = a_2 \qquad \Rightarrow \qquad a_2(a_1 - 1) = 0 \qquad \qquad \text{(2)}$

From (1), $a_2 = 0$ , which satisfies (2) for any $a_1$ . So the reflexive polynomials of degree 2 are:

$\boxed{x^2 - a_1 x \quad \text{for any real } a_1}$

Verification: $x^2 - a_1 x = x(x - a_1)$ has roots $0$ and $a_1$ , which are indeed $a_2 = 0$ and $a_1$ . ✓

Degree 3: $\mathrm{P}(x) = x^3 - a_1 x^2 + a_2 x - a_3$ . By Vieta’s formulas:

$a_1 + a_2 + a_3 = a_1 \qquad \Rightarrow \qquad a_2 + a_3 = 0 \qquad \Rightarrow \qquad a_3 = -a_2 \qquad \qquad \text{(3)}$

$a_1 a_2 + a_2 a_3 + a_3 a_1 = a_2 \qquad \qquad \text{(4)}$

$a_1 a_2 a_3 = a_3 \qquad \Rightarrow \qquad a_3(a_1 a_2 - 1) = 0 \qquad \qquad \text{(5)}$

From (5), either $a_3 = 0$ or $a_1 a_2 = 1$ .

Subcase $a_3 = 0$ : From (3), $a_2 = 0$ . Equation (4) becomes $0 = 0$ . So $a_1$ is free, giving the polynomial $x^3 - a_1 x^2$ .

Verification: $x^3 - a_1 x^2 = x^2(x - a_1)$ has roots $0, 0, a_1$ , which are $a_2 = 0$ , $a_3 = 0$ , $a_1$ . ✓

Subcase $a_1 a_2 = 1$ , $a_3 \neq 0$ : From (3), $a_3 = -a_2$ . Substitute into (4):

$a_1 a_2 + a_2(-a_2) + (-a_2) a_1 = a_2$

$a_1 a_2 - a_2^2 - a_1 a_2 = a_2$

$-a_2^2 = a_2$

$a_2(a_2 + 1) = 0$

Since $a_3 = -a_2 \neq 0$ , we have $a_2 \neq 0$ , so $a_2 = -1$ . Then $a_3 = -(-1) = 1$ and $a_1 = \frac{1}{a_2} = -1$ .

Verification: $\mathrm{P}(x) = x^3 + x^2 - x - 1 = (x+1)^2(x-1)$ , with roots $-1, -1, 1$ . These are $a_1 = -1$ , $a_2 = -1$ , $a_3 = 1$ . ✓

So the reflexive polynomials of degree $\leqslant 3$ are:

$\boxed{x - a_1, \quad x^2 - a_1 x, \quad x^3 - a_1 x^2 \quad (\text{any real } a_1), \qquad x^3 + x^2 - x - 1}$

Part (ii)

For a reflexive polynomial of degree $n$ , the roots are $a_1, a_2, \ldots, a_n$ . By Vieta’s formulas:

$a_1 = \sum_{r=1}^{n} a_r \qquad \Rightarrow \qquad \sum_{r=2}^{n} a_r = 0 \qquad \qquad \text{(6)}$

$a_2 = \sum_{1 \leqslant i < j \leqslant n} a_i a_j \qquad \qquad \text{(7)}$

Consider the square of $\sum_{r=2}^{n} a_r$ :

$\left(\sum_{r=2}^{n} a_r\right)^2 = \sum_{r=2}^{n} a_r^2 + 2 \sum_{2 \leqslant i < j \leqslant n} a_i a_j$

From (6), the left side is $0$ , so:

$0 = \sum_{r=2}^{n} a_r^2 + 2 \sum_{2 \leqslant i < j \leqslant n} a_i a_j \qquad \qquad \text{(8)}$

Now we relate $\sum_{2 \leqslant i < j \leqslant n} a_i a_j$ to $a_2$ . From (7):

$a_2 = \sum_{1 \leqslant i < j \leqslant n} a_i a_j = a_1 \sum_{r=2}^{n} a_r + \sum_{2 \leqslant i < j \leqslant n} a_i a_j$

By (6), $a_1 \sum_{r=2}^{n} a_r = a_1 \cdot 0 = 0$ , so:

$a_2 = \sum_{2 \leqslant i < j \leqslant n} a_i a_j$

Substituting into (8):

$0 = \sum_{r=2}^{n} a_r^2 + 2a_2$

$2a_2 = -\sum_{r=2}^{n} a_r^2 = -a_2^2 - a_3^2 - \cdots - a_n^2$

$\boxed{2a_2 = -a_2^2 - a_3^2 - \cdots - a_n^2} \qquad \blacksquare$

Deduction: Rearranging:

$a_2^2 + 2a_2 + a_3^2 + \cdots + a_n^2 = 0$

$(a_2 + 1)^2 + a_3^2 + \cdots + a_n^2 = 1$

Since squares are non-negative:

$a_3^2 + a_4^2 + \cdots + a_n^2 = 1 - (a_2 + 1)^2 \leqslant 1$

If all coefficients are integers and $a_n \neq 0$ , then $a_n^2 \geqslant 1$ . Combined with $a_3^2 + \cdots + a_n^2 \leqslant 1$ , we must have $a_n^2 = 1$ and $a_3 = a_4 = \cdots = a_{n-1} = 0$ .

From (6): $a_2 + a_n = 0$ (since $a_3 = \cdots = a_{n-1} = 0$ ), so $a_2 = -a_n = \mp 1$ .

From $a_1 a_2 \cdots a_n = a_n$ (the product of roots equals $a_n$ ), dividing by $a_n \neq 0$ :

$a_1 a_2 \cdots a_{n-1} = 1$

Since $a_3 = \cdots = a_{n-1} = 0$ (and $n > 3$ means at least one such coefficient exists), the left side is $0$ . But the right side is $1$ . Contradiction.

Therefore, if all coefficients are integers and $a_n \neq 0$ , then $n \leqslant 3$ . $\qquad \blacksquare$

Part (iii)

From part (ii), any reflexive polynomial with integer coefficients and $n > 3$ must have $a_n = 0$ . Then:

$\mathrm{P}(x) = x^n - a_1 x^{n-1} + \cdots + (-1)^{n-1} a_{n-1} x = x \cdot \mathrm{Q}(x)$

where $\mathrm{Q}(x) = x^{n-1} - a_1 x^{n-2} + \cdots + (-1)^{n-1} a_{n-1}$ .

Since $\mathrm{P}(x) = 0$ has root $x = 0 = a_n$ , and the remaining roots are $a_1, \ldots, a_{n-1}$ , the polynomial $\mathrm{Q}(x)$ must have roots $a_1, \ldots, a_{n-1}$ . Moreover, $\mathrm{Q}(x)$ has the correct sign alternation (the coefficient of $x^{n-1-k}$ is $(-1)^k a_k$ ). So $\mathrm{Q}(x)$ is itself reflexive of degree $n-1$ .

By induction, we keep factoring out $x$ until we reach a reflexive polynomial of degree $\leqslant 3$ with integer coefficients. From part (i), these are:

Degree 1: $x - a_1$ with $a_1 \in \mathbb{Z}$
Degree 2: $x^2 - a_1 x$ with $a_1 \in \mathbb{Z}$
Degree 3: $x^3 - a_1 x^2$ with $a_1 \in \mathbb{Z}$ , or $x^3 + x^2 - x - 1 = (x+1)^2(x-1)$

But $x^2 - a_1 x = x(x - a_1)$ is just $x$ times a degree-1 reflexive, and $x^3 - a_1 x^2 = x^2(x - a_1)$ is $x^2$ times a degree-1 reflexive. So the complete list of reflexive polynomials with integer coefficients is:

$\boxed{(x - a_1) \cdot x^r \qquad \text{or} \qquad (x+1)^2(x-1) \cdot x^r \qquad \text{with } a_1 \in \mathbb{Z},\ r = 0, 1, 2, \ldots}$

Equivalently: $x^r(x - a_1)$ for any integer $a_1$ and non-negative integer $r$ , or $x^r(x+1)^2(x-1)$ for any non-negative integer $r$ . $\qquad \blacksquare$

Examiner Notes

此题概念新颖,需仔细理解定义中”系数=根”的含义。第(ii)部分利用平方和>=0推出a2=0,进而限制系数取值。整系数条件下n<=3的证明是关键难点。

Question 5

Topic: 纯数 | Difficulty: Hard | Marks: 20

Problem

5 (i) Let $f(x) = \frac{x}{\sqrt{x^2 + p}},$ where $p$ is a non-zero constant. Sketch the curve $y = f(x)$ for $x \geqslant 0$ in the case $p > 0$ .

(ii) Let $I = \int \frac{1}{(b^2 - y^2)\sqrt{c^2 - y^2}} \, dy,$ where $b$ and $c$ are positive constants. Use the substitution $y = \frac{bx}{\sqrt{x^2 + p}}$ , where $p$ is a suitably chosen constant, to show that $I = \int \frac{1}{b^2 + (b^2 - c^2)x^2} \, dx.$ Evaluate $\int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \, dy.$ [ Note: $\int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \frac{x}{a} + \text{ constant.}$ ]

Hence evaluate $\int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$

(iii) By means of a suitable substitution, evaluate $\int_{\frac{1}{\sqrt{2}}}^1 \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy.$

Hint

(i) $f(x) = \frac{x}{\sqrt{x^2 + p}}$ $f'(x) = \frac{\sqrt{x^2 + p} - x \frac{1}{2} \frac{2x}{\sqrt{x^2 + p}}}{x^2 + p} = \frac{p}{(x^2 + p)^{\frac{3}{2}}}$

[3]

(ii) for answers using substitution from question paper

$y = \frac{bx}{\sqrt{x^2 + p}} \Rightarrow \frac{dy}{dx} = \frac{bp}{(x^2 + p)^{\frac{3}{2}}}$ $b^2 - y^2 = b^2 - \frac{b^2x^2}{x^2 + p} = \frac{(b^2 - b^2)x^2 + b^2p}{x^2 + p} = \frac{b^2p}{x^2 + p}$

$c^2 - y^2 = c^2 - \frac{b^2x^2}{x^2 + p} = \frac{(c^2 - b^2)x^2 + c^2p}{x^2 + p}$

So $\int \frac{1}{(b^2 - y^2)\sqrt{c^2 - y^2}} \, dy = \int \frac{(x^2 + p)}{b^2p} \frac{\sqrt{x^2 + p}}{\sqrt{[(c^2 - b^2)x^2 + c^2p]}} \frac{bp}{(x^2 + p)^{\frac{3}{2}}} \, dx$

$= \int \frac{1}{b\sqrt{[(c^2 - b^2)x^2 + c^2p]}} \, dx$

M1 [4]

Let $c^2 = 2$ M1

Let $b^2 = 3$ M1

Thus $\int_1^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} \, dy = \int_?^? \frac{1}{3 + x^2} \, dx = \left[ \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) \right]_?^?$ M1 M1 M1 [5]

Let $y = \frac{1}{x}$ , $\frac{dy}{dx} = -\frac{1}{x^2}$ M1

$\int_{\frac{1}{\sqrt{2}}}^1 \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} \, dy = \int_{\sqrt{2}}^1 \frac{1}{x} \frac{1}{\frac{3}{x^2} - 1} \frac{1}{\sqrt{\frac{2}{x^2} - 1}} \times -\frac{1}{x^2} \, dx = \int_1^{\sqrt{2}} \frac{1}{(3 - x^2)\sqrt{2 - x^2}} \, dx$ M1

and so is same answer as previous part A1 ft [3]

(ii) for answers using correct substitution where candidates have realised there is a misprint

$y = \frac{cx}{\sqrt{x^2 + p}} \Rightarrow \frac{dy}{dx} = \frac{cp}{(x^2 + p)^{\frac{3}{2}}}$

$b^2 - y^2 = b^2 - \frac{c^2x^2}{x^2 + p} = \frac{(b^2 - c^2)x^2 + b^2p}{x^2 + p}$

$c^2 - y^2 = c^2 - \frac{c^2x^2}{x^2 + p} = \frac{c^2p}{x^2 + p}$

Choose $p = 1$ B1

So $\int \frac{1}{(b^2 - y^2)\sqrt{c^2 - y^2}} \, dy = \int \frac{x^2 + 1}{b^2 + (b^2 - c^2)x^2} \frac{\sqrt{x^2 + 1}}{c} \frac{c}{(x^2 + 1)^{\frac{3}{2}}} \, dx$ M1

$= \int \frac{1}{b^2 + (b^2 - c^2)x^2} \, dx$ A1* [4]

Let $c^2 = 2$ Then $(x^2 + 1)y^2 = 2x^2$ so $x^2 = \frac{y^2}{2-y^2}$ and when $y = 1, x = 1$

and $y \rightarrow \sqrt{2}, x \rightarrow \infty$

Let $b^2 = 3$ [B1 M1 A1]

Thus

$\int_{1}^{\sqrt{2}} \frac{1}{(3 - y^2)\sqrt{2 - y^2}} dy = \int_{1}^{\infty} \frac{1}{3 + x^2} dx = \left[ \frac{1}{\sqrt{3}} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) \right]_{1}^{\infty} = \frac{1}{\sqrt{3}} \left( \frac{\pi}{2} - \frac{\pi}{6} \right) = \frac{\pi}{3\sqrt{3}}$

[M1 A1] [5]

Let $y = \frac{1}{x}, \frac{dy}{dx} = -\frac{1}{x^2}$ [M1]

$\int_{\frac{1}{\sqrt{2}}}^{1} \frac{y}{(3y^2 - 1)\sqrt{2y^2 - 1}} dy = \int_{\sqrt{2}}^{1} \frac{\frac{1}{x}}{\frac{3}{x^2} - 1} \frac{1}{\sqrt{\frac{2}{x^2} - 1}} \times -\frac{1}{x^2} dx = \int_{1}^{\sqrt{2}} \frac{1}{(3 - x^2)\sqrt{2 - x^2}} dx$

[M1]

and so is $\frac{\pi}{3\sqrt{3}}$ [A1 ft] [3]

(iii) If

$y = \frac{bx}{\sqrt{x^2 + p}} \Rightarrow \frac{dy}{dx} = \frac{bp}{(x^2 + p)^{\frac{3}{2}}}$

and so

$(x^2 + p)y^2 = b^2x^2, x^2 = \frac{py^2}{b^2 - y^2}$ [M1]

thus

$\int \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} dy = \int \frac{1}{\left( \frac{3b^2x^2}{x^2 + p} - 1 \right) \sqrt{\frac{2b^2x^2}{x^2 + p} - 1}} \frac{bp}{(x^2 + p)^{\frac{3}{2}}} dx$

$= \int \frac{bp}{(3b^2x^2 - (x^2 + p)) \sqrt{2b^2x^2 - (x^2 + p)}} dx$

[M1]

Choosing $2b^2 = 1$ and $p = -1$ we have [B1]

$\int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{(3y^2 - 1)\sqrt{2y^2 - 1}} dy = \int_{\infty}^{\sqrt{2}} \frac{-\frac{1}{\sqrt{2}}}{\frac{1}{2}x^2 + 1} dx = \sqrt{2} \int_{\sqrt{2}}^{\infty} \frac{1}{x^2 + 2} dx$

$= \sqrt{2} \left[ \frac{1}{\sqrt{2}} \tan^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_{\sqrt{2}}^{\infty} = \frac{\pi}{4}$

A1 [5]

Model Solution

Part (i)

We have $f(x) = \dfrac{x}{\sqrt{x^2 + p}}$ with $p > 0$ .

At $x = 0$ : $f(0) = 0$ .

As $x \to \infty$ : $f(x) = \dfrac{x}{\sqrt{x^2+p}} = \dfrac{1}{\sqrt{1+p/x^2}} \to 1^-$ .

Differentiating: $f'(x) = \frac{\sqrt{x^2+p} - x \cdot \dfrac{x}{\sqrt{x^2+p}}}{x^2+p} = \frac{(x^2+p) - x^2}{(x^2+p)^{3/2}} = \frac{p}{(x^2+p)^{3/2}}$

Since $p > 0$ , $f'(x) > 0$ for all $x \geqslant 0$ , so $f$ is strictly increasing.

We can also note that for $x > 0$ : $f''(x) = \frac{-3px}{(x^2+p)^{5/2}} < 0$

so the curve is concave (curving towards the asymptote from below).

The sketch shows the curve passing through the origin, increasing monotonically, and approaching the horizontal asymptote $y = 1$ from below. The curve is concave throughout $x > 0$ .

Part (ii)

Note: The question paper contains a misprint. The substitution should be $y = \dfrac{cx}{\sqrt{x^2+p}}$ (with $c$ not $b$ ). We use the corrected substitution.

We use $y = \dfrac{cx}{\sqrt{x^2+p}}$ where $p$ is to be determined.

Computing $\dfrac{dy}{dx}$ : $\frac{dy}{dx} = \frac{c\sqrt{x^2+p} - cx \cdot \dfrac{x}{\sqrt{x^2+p}}}{x^2+p} = \frac{c(x^2+p) - cx^2}{(x^2+p)^{3/2}} = \frac{cp}{(x^2+p)^{3/2}}$

Computing $b^2 - y^2$ : $b^2 - y^2 = b^2 - \frac{c^2x^2}{x^2+p} = \frac{b^2(x^2+p) - c^2x^2}{x^2+p} = \frac{b^2p + (b^2-c^2)x^2}{x^2+p}$

Computing $c^2 - y^2$ : $c^2 - y^2 = c^2 - \frac{c^2x^2}{x^2+p} = \frac{c^2(x^2+p) - c^2x^2}{x^2+p} = \frac{c^2p}{x^2+p}$

So $\sqrt{c^2-y^2} = \dfrac{c\sqrt{p}}{\sqrt{x^2+p}}$ .

Substituting into the integral: $I = \int \frac{1}{(b^2-y^2)\sqrt{c^2-y^2}} \, dy = \int \frac{x^2+p}{b^2p+(b^2-c^2)x^2} \cdot \frac{\sqrt{x^2+p}}{c\sqrt{p}} \cdot \frac{cp}{(x^2+p)^{3/2}} \, dx$

Simplifying the factors: $I = \int \frac{cp \cdot (x^2+p)^{3/2}}{[b^2p+(b^2-c^2)x^2] \cdot c\sqrt{p} \cdot (x^2+p)^{3/2}} \, dx = \int \frac{\sqrt{p}}{b^2p + (b^2-c^2)x^2} \, dx$

For this to equal $\displaystyle\int \frac{1}{b^2 + (b^2-c^2)x^2} \, dx$ , we choose $p = 1$ :

$I = \int \frac{1}{b^2 + (b^2-c^2)x^2} \, dx$

as required. $\qquad \blacksquare$

Evaluating the specific integral:

With $b^2 = 3$ and $c^2 = 2$ , we have $b^2 - c^2 = 1$ , and the substitution is $y = \dfrac{\sqrt{2}\,x}{\sqrt{x^2+1}}$ .

From this: $y^2(x^2+1) = 2x^2$ , giving $x^2 = \dfrac{y^2}{2-y^2}$ .

When $y = 1$ : $x^2 = 1$ , so $x = 1$ .

When $y = \sqrt{2}$ : $x^2 \to \infty$ , so $x \to \infty$ .

Therefore: $\int_1^{\sqrt{2}} \frac{1}{(3-y^2)\sqrt{2-y^2}} \, dy = \int_1^{\infty} \frac{1}{3+x^2} \, dx$

Using the given formula with $a = \sqrt{3}$ : $= \left[\frac{1}{\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}\right]_1^{\infty} = \frac{1}{\sqrt{3}}\left(\frac{\pi}{2} - \tan^{-1}\frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}}\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{3} = \frac{\pi}{3\sqrt{3}}$

Hence evaluating the second integral:

Consider the substitution $y = \dfrac{1}{x}$ , so $dy = -\dfrac{1}{x^2}\,dx$ .

When $y = \dfrac{1}{\sqrt{2}}$ : $x = \sqrt{2}$ . When $y = 1$ : $x = 1$ .

Transforming the integrand: $\frac{y}{(3y^2-1)\sqrt{2y^2-1}} = \frac{1/x}{(3/x^2-1)\sqrt{2/x^2-1}}$

$= \frac{1/x}{\dfrac{3-x^2}{x^2} \cdot \dfrac{\sqrt{2-x^2}}{x}} = \frac{1/x \cdot x^3}{(3-x^2)\sqrt{2-x^2}} = \frac{x^2}{(3-x^2)\sqrt{2-x^2}}$

So: $\int_{1/\sqrt{2}}^1 \frac{y}{(3y^2-1)\sqrt{2y^2-1}} \, dy = \int_{\sqrt{2}}^1 \frac{x^2}{(3-x^2)\sqrt{2-x^2}} \cdot \left(-\frac{1}{x^2}\right) dx = \int_1^{\sqrt{2}} \frac{1}{(3-x^2)\sqrt{2-x^2}} \, dx$

This is identical to the integral evaluated above, so:

$\int_{1/\sqrt{2}}^1 \frac{y}{(3y^2-1)\sqrt{2y^2-1}} \, dy = \frac{\pi}{3\sqrt{3}}$

Part (iii)

We evaluate $\displaystyle\int_{1/\sqrt{2}}^1 \frac{1}{(3y^2-1)\sqrt{2y^2-1}} \, dy$ .

The integrand involves $y^2$ throughout (no $y$ in the numerator), so the substitution $y = 1/x$ from part (ii) does not reduce it to the same form. Instead we use $y = \dfrac{bx}{\sqrt{x^2+p}}$ with suitably chosen $b$ and $p$ .

With $y = \dfrac{bx}{\sqrt{x^2+p}}$ : $y^2 = \frac{b^2x^2}{x^2+p}$

$3y^2 - 1 = \frac{3b^2x^2 - (x^2+p)}{x^2+p} = \frac{(3b^2-1)x^2 - p}{x^2+p}$

$2y^2 - 1 = \frac{2b^2x^2 - (x^2+p)}{x^2+p} = \frac{(2b^2-1)x^2 - p}{x^2+p}$

$\frac{dy}{dx} = \frac{bp}{(x^2+p)^{3/2}}$

The integral becomes: $\int \frac{x^2+p}{(3b^2-1)x^2 - p} \cdot \frac{\sqrt{x^2+p}}{\sqrt{(2b^2-1)x^2-p}} \cdot \frac{bp}{(x^2+p)^{3/2}} \, dx = \int \frac{bp}{[(3b^2-1)x^2 - p]\sqrt{(2b^2-1)x^2 - p}} \, dx$

We choose $2b^2 - 1 = 0$ , i.e., $b^2 = \dfrac{1}{2}$ , $b = \dfrac{1}{\sqrt{2}}$ , and $p = -1$ .

Then:

$bp = \dfrac{1}{\sqrt{2}} \cdot (-1) = -\dfrac{1}{\sqrt{2}}$
$(3b^2-1)x^2 - p = \dfrac{1}{2}x^2 + 1 = \dfrac{x^2+2}{2}$
$(2b^2-1)x^2 - p = 0 + 1 = 1$

The integral simplifies to: $\int \frac{-1/\sqrt{2}}{\dfrac{x^2+2}{2} \cdot 1} \, dx = \int \frac{-\sqrt{2}}{x^2+2} \, dx$

Limits: From $y^2 = \dfrac{x^2}{2(x^2-1)}$ , we get $x^2 = \dfrac{2y^2}{2y^2-1}$ .

When $y = 1$ : $x^2 = 2$ , so $x = \sqrt{2}$ .

When $y = \dfrac{1}{\sqrt{2}}$ : $y^2 = \dfrac{1}{2}$ , $x^2 = \dfrac{1}{0^+} \to \infty$ .

Therefore: $\int_{1/\sqrt{2}}^1 \frac{1}{(3y^2-1)\sqrt{2y^2-1}} \, dy = \int_{\infty}^{\sqrt{2}} \frac{-\sqrt{2}}{x^2+2} \, dx = \sqrt{2}\int_{\sqrt{2}}^{\infty} \frac{1}{x^2+2} \, dx$

Using the formula with $a = \sqrt{2}$ : $= \sqrt{2}\left[\frac{1}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}}\right]_{\sqrt{2}}^{\infty} = \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\frac{\pi}{2} - \tan^{-1} 1\right) = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

Examiner Notes

注意试卷勘误:第(ii)部分代换中b应改为c。选择合适的p值是解题关键;需正确处理积分限的变换。

Question 6

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

6 The point $P$ in the Argand diagram is represented by the complex number $z$ , which satisfies $zz^* - az^* - a^*z + aa^* - r^2 = 0.$ Here, $r$ is a positive real number and $r^2 \neq a^*a$ . By writing $|z - a|^2$ as $(z - a)(z - a)^*$ , show that the locus of $P$ is a circle, $C$ , the radius and the centre of which you should give.

(i) The point $Q$ is represented by $\omega$ , and is related to $P$ by $\omega = \frac{1}{z}$ . Let $C'$ be the locus of $Q$ . Show that $C'$ is also a circle, and give its radius and centre.

If $C$ and $C'$ are the same circle, show that $(|a|^2 - r^2)^2 = 1$ and that either $a$ is real or $a$ is imaginary. Give sketches to indicate the position of $C$ in these two cases.

(ii) Suppose instead that the point $Q$ is represented by $\omega$ , where $\omega = \frac{1}{z^*}$ . If the locus of $Q$ is $C$ , is it the case that either $a$ is real or $a$ is imaginary?

Hint

$|z - a|^2 = (z - a)(z - a)^* = (z - a)(z^* - a^*) = zz^* - az^* - a^*z + aa^*$

So $z$ satisfies $|z - a|^2 = r^2$ which means that the locus of P is a circle (C) centre $a$ and radius $r$ (which does not pass through the origin as $r^2 \neq aa^*$ .) A1 [2]

(i) As $w = \frac{1}{z}$ , $z = \frac{1}{w}$ so $\frac{1}{w} \frac{1}{w^*} - a \frac{1}{w^*} - a^* \frac{1}{w} + aa^* - r^2 = 0$

$(aa^* - r^2)ww^* - aw - a^*w^* + 1 = 0$

$ww^* - \frac{a}{aa^* - r^2}w - \frac{a^*}{aa^* - r^2}w^* + \frac{1}{aa^* - r^2} = 0$

$ww^* - \left(\frac{a^*}{aa^* - r^2}\right)^* w - \left(\frac{a^*}{aa^* - r^2}\right)w^* + \left(\frac{a^*}{aa^* - r^2}\right)\left(\frac{a}{aa^* - r^2}\right)$ $= \left(\frac{a^*}{aa^* - r^2}\right)\left(\frac{a}{aa^* - r^2}\right) - \frac{1}{aa^* - r^2}$

$\left|w - \frac{a^*}{aa^* - r^2}\right|^2 = \frac{aa^*}{(aa^* - r^2)^2} - \frac{1}{aa^* - r^2} = \frac{r^2}{(aa^* - r^2)^2}$

So C’ is a circle centre $\frac{a^*}{aa^* - r^2}$ with radius $\left|\frac{r}{aa^* - r^2}\right|$ A1 [3]

If C and C’ are the same circle, then

$a = \frac{a^*}{aa^* - r^2}$

and

$r^2 = \left|\frac{r}{aa^* - r^2}\right|^2$

Thus

$r^2(|a|^2 - r^2)^2 = r^2$

and dividing by $r^2 \neq 0$ , $(|a|^2 - r^2)^2 = 1$ as required. A1* [3]

So $|a|^2 - r^2 = \pm 1$ and the equation equating the centres becomes $a = \pm a^*$ M1

If $a = c + di$ , and $a = a^*$ , $c + di = c - di$ implying $d = 0$ and hence $a$ is real. M1 A1

If $a = c + di$ , and $a = -a^*$ , $c + di = -c + di$ implying $c = 0$ and hence $a$ is imaginary. M1 A1

G1 [7]

(ii) If $w = \frac{1}{z^*}$ then

$\left| w^* - \frac{a^*}{aa^* - r^2} \right|^2 = \frac{r^2}{(aa^* - r^2)^2}$

$\left| w - \frac{a}{aa^* - r^2} \right|^2 = \frac{r^2}{(aa^* - r^2)^2}$

and thus, as before $|a|^2 - r^2 = \pm 1$ but now $a = \frac{a}{aa^* - r^2}$ A1

So in the case $|a|^2 - r^2 = +1$ , then any $a$ with $|a| = \sqrt{r^2 + 1}$ is possible; in the case $|a|^2 - r^2 = -1$ , $a = 0$ (in which case $r = 1$ ) A1

So, it is not the case that $a$ is either real or imaginary. A1 [5]

Model Solution

Main result:

Recall that $|z-a|^2 = (z-a)(z-a)^* = (z-a)(z^*-a^*) = zz^* - az^* - a^*z + aa^*$ .

The given equation $zz^* - az^* - a^*z + aa^* - r^2 = 0$ is therefore equivalent to:

$|z-a|^2 = r^2$

This is the equation of a circle $C$ with centre $a$ and radius $r$ .

Since $r^2 \neq aa^* = |a|^2$ , the circle does not pass through the origin. $\qquad \blacksquare$

Part (i)

Since $\omega = 1/z$ , we have $z = 1/\omega$ and $z^* = 1/\omega^*$ . Substituting into the equation of $C$ :

$\frac{1}{\omega} \cdot \frac{1}{\omega^*} - a \cdot \frac{1}{\omega^*} - a^* \cdot \frac{1}{\omega} + aa^* - r^2 = 0$

Multiplying through by $\omega\omega^*$ :

$1 - a\omega - a^*\omega^* + (aa^* - r^2)\omega\omega^* = 0$

Since $r^2 \neq aa^*$ , we may divide by $(aa^* - r^2)$ :

$\omega\omega^* - \frac{a}{aa^* - r^2}\omega - \frac{a^*}{aa^* - r^2}\omega^* + \frac{1}{aa^* - r^2} = 0$

Note that $\dfrac{a}{aa^* - r^2} = \left(\dfrac{a^*}{aa^* - r^2}\right)^*$ since $aa^* - r^2$ is real.

Completing the square: $\left|\omega - \frac{a^*}{aa^* - r^2}\right|^2 = \left|\frac{a^*}{aa^* - r^2}\right|^2 - \frac{1}{aa^* - r^2} = \frac{aa^*}{(aa^* - r^2)^2} - \frac{1}{aa^* - r^2}$

$= \frac{aa^* - (aa^* - r^2)}{(aa^* - r^2)^2} = \frac{r^2}{(aa^* - r^2)^2}$

So $C'$ is a circle with centre $\dfrac{a^*}{aa^* - r^2}$ and radius $\dfrac{r}{|aa^* - r^2|}$ . $\qquad \blacksquare$

If $C$ and $C'$ are the same circle:

The centres must be equal: $a = \frac{a^*}{aa^* - r^2} \qquad \text{...(1)}$

The radii must be equal: $r^2 = \frac{r^2}{(aa^* - r^2)^2} \qquad \text{...(2)}$

From (2), since $r \neq 0$ , we divide by $r^2$ : $(aa^* - r^2)^2 = 1 \implies (|a|^2 - r^2)^2 = 1$

So $|a|^2 - r^2 = \pm 1$ .

Case 1: $|a|^2 - r^2 = 1$

From (1): $a = \dfrac{a^*}{1} = a^*$ .

Writing $a = c + di$ : $c + di = c - di$ , so $d = 0$ and $a$ is real.

Case 2: $|a|^2 - r^2 = -1$

From (1): $a = \dfrac{a^*}{-1} = -a^*$ .

Writing $a = c + di$ : $c + di = -c + di$ , so $c = 0$ and $a$ is imaginary.

Sketches:

When $a$ is real (Case 1): $|a|^2 = r^2 + 1$ , so $|a| > r$ . The circle $C$ is centred on the real axis at $a$ with radius $r < |a|$ , so it does not enclose the origin.
When $a$ is imaginary (Case 2): $|a|^2 = r^2 - 1$ , so $r > |a|$ . The circle $C$ is centred on the imaginary axis at $a$ with radius $r > |a|$ , so it encloses the origin.

Part (ii)

Now $\omega = 1/z^*$ , so $z^* = 1/\omega$ and $z = 1/\omega^*$ . Substituting:

$\frac{1}{\omega^*} \cdot \frac{1}{\omega} - a \cdot \frac{1}{\omega} - a^* \cdot \frac{1}{\omega^*} + aa^* - r^2 = 0$

Multiplying by $\omega\omega^*$ : $1 - a\omega^* - a^*\omega + (aa^* - r^2)\omega\omega^* = 0$

Dividing by $(aa^* - r^2)$ : $\omega\omega^* - \frac{a^*}{aa^* - r^2}\omega - \frac{a}{aa^* - r^2}\omega^* + \frac{1}{aa^* - r^2} = 0$

Completing the square: $\left|\omega - \frac{a}{aa^* - r^2}\right|^2 = \frac{|a|^2}{(aa^* - r^2)^2} - \frac{1}{aa^* - r^2} = \frac{|a|^2 - (aa^* - r^2)}{(aa^* - r^2)^2} = \frac{r^2}{(aa^* - r^2)^2}$

So the locus of $Q$ is a circle with centre $\dfrac{a}{|a|^2 - r^2}$ and radius $\dfrac{r}{\bigl||a|^2 - r^2\bigr|}$ .

If this is to be the circle $C$ :

Centre: $a = \dfrac{a}{|a|^2 - r^2}$ , which gives $|a|^2 - r^2 = 1$ (since $a \neq 0$ ).

Radius: $r^2 = \dfrac{r^2}{(|a|^2 - r^2)^2}$ , which gives $(|a|^2 - r^2)^2 = 1$ .

These are consistent: $|a|^2 - r^2 = 1$ satisfies $(|a|^2 - r^2)^2 = 1$ .

But the centre condition $a = \dfrac{a}{|a|^2 - r^2}$ reduces to $|a|^2 - r^2 = 1$ , which places no constraint on the argument of $a$ . Any complex number with $|a|^2 = r^2 + 1$ works.

For example, $a = 1 + i$ with $r = 1$ gives $|a|^2 = 2 = r^2 + 1$ . The circle $C$ has centre $1 + i$ and radius $1$ . Since $|a|^2 - r^2 = 1$ , the transformed circle $C'$ also has centre $\dfrac{1+i}{1} = 1 + i$ and radius $\dfrac{1}{1} = 1$ , so $C' = C$ . But $a = 1 + i$ is neither real nor imaginary.

Therefore, it is not the case that either $a$ is real or $a$ is imaginary. $\qquad \blacksquare$

Examiner Notes

需注意条件r^2!=aa的几何含义;反演变换将圆映射为圆(或直线)是经典结论;第(ii)部分omega=1/z的情况结论与(i)不同,需独立分析。

Question 7

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

7 The Devil’s Curve is given by $y^2(y^2 - b^2) = x^2(x^2 - a^2),$ where $a$ and $b$ are positive constants.

(i) In the case $a = b$ , sketch the Devil’s Curve.

(ii) Now consider the case $a = 2$ and $b = \sqrt{5}$ , and $x \geqslant 0, y \geqslant 0$ .

(a) Show by considering a quadratic equation in $x^2$ that either $0 \leqslant y \leqslant 1$ or $y \geqslant 2$ .

(b) Describe the curve very close to and very far from the origin.

(c) Find the points at which the tangent to the curve is parallel to the $x$ -axis and the point at which the tangent to the curve is parallel to the $y$ -axis.

Sketch the Devil’s Curve in this case.

(iii) Sketch the Devil’s Curve in the case $a = 2$ and $b = \sqrt{5}$ again, but with $-\infty < x < \infty$ and $-\infty < y < \infty$ .

Hint

(i) $y^2(y^2 - a^2) = x^2(x^2 - a^2)$ $y^4 - a^2y^2 + \frac{a^4}{4} = x^4 - a^2x^2 + \frac{a^4}{4}$

$\left(y^2 - \frac{a^2}{2}\right)^2 = \left(x^2 - \frac{a^2}{2}\right)^2$ $y^2 - \frac{a^2}{2} = \pm \left(x^2 - \frac{a^2}{2}\right)$

So $x^2 = y^2$ which gives $y = \pm x$ or $x^2 + y^2 = a^2$

Alternatively

$y^4 - x^4 = a^2y^2 - a^2x^2$ so $(y^2 - x^2)(y^2 + x^2) - a^2(y^2 - x^2) = 0$

G1 [3]

(ii) $y^2(y^2 - 5) = x^2(x^2 - 4)$

a) $(x^2)^2 - 4x^2 - y^2(y^2 - 5) = 0$

So for real $x^2$ , $16 + 4y^2(y^2 - 5) \geq 0$

$(y^2)^2 - 5y^2 + 4 \ge 0$ $(y^2 - 1)(y^2 - 4) \ge 0$ $(y - 1)(y + 1)(y - 2)(y + 2) \ge 0$

As $y \ge 0$ ,

$(y - 1)(y - 2) \ge 0$

So $0 \le y \le 1$ or $y \ge 2$ A1* [3]

b) For small x and y,

$y^4 - 5y^2 = x^4 - 4x^2$ becomes $5y^2 \approx 4x^2$ so $y \approx \pm \frac{2x}{\sqrt{5}}$ B1

For large x and y,

$y^4 - 5y^2 = x^4 - 4x^2$ becomes $y^4 \approx x^4$ so $y \approx \pm x$ B1 [2]

$y^2(y^2 - 5) = x^2(x^2 - 4)$

$(4y^3 - 10y) \frac{dy}{dx} = 4x^3 - 8x$

$\frac{dy}{dx} = 0 \Rightarrow 4x^3 - 8x = 0$

$4x(x^2 - 2) = 0$

So $x = 0$ , $y = 0, \sqrt{5}$ or $x = \sqrt{2}$ , $y = 1, 2$

Thus $(0, \sqrt{5}), (\sqrt{2}, 1), (\sqrt{2}, 2)$ but not $(0, 0)$ A1 B1

$\frac{dx}{dy} = 0 \Rightarrow 4y^3 - 10y = 0$ $2y(2y^2 - 5) = 0$

So $y = 0$ , $x = 0, 2$ but $y = \sqrt{\frac{5}{2}}$ gives $x$ complex E1

Thus $(2, 0)$ but not $(0, 0)$ A1 [7]

Sketch key points: $(0,0)$ , $(\sqrt{2},\sqrt{2})$ , and $(2,0)$ . The point with $y=1$ should be included on the appropriate branch.

[G1 G1 G1 G1] [4]

(iii) Sketch the transformed curve, indicating the same intercepts and the corresponding branch behaviour clearly. [G1 ft] [1]

Model Solution

Part (i)

When $a = b$ , the equation becomes

$y^2(y^2 - a^2) = x^2(x^2 - a^2).$

Expanding both sides:

$y^4 - a^2 y^2 = x^4 - a^2 x^2.$

Rearranging:

$y^4 - x^4 = a^2(y^2 - x^2).$

Factorising the left side as a difference of squares:

$(y^2 - x^2)(y^2 + x^2) = a^2(y^2 - x^2).$

$(y^2 - x^2)\bigl(y^2 + x^2 - a^2\bigr) = 0.$

This gives two factors:

$y^2 - x^2 = 0$ , i.e.\ $y = \pm x$ (a pair of straight lines through the origin), or
$x^2 + y^2 = a^2$ (a circle of radius $a$ centred at the origin).

So the Devil’s Curve with $a = b$ consists of the circle $x^2 + y^2 = a^2$ together with the lines $y = x$ and $y = -x$ .

Part (ii)(a)

With $a = 2$ and $b = \sqrt{5}$ , the equation is

$y^2(y^2 - 5) = x^2(x^2 - 4).$

Expanding:

$y^4 - 5y^2 = x^4 - 4x^2.$

Rearranging as a quadratic in $x^2$ :

$(x^2)^2 - 4x^2 - y^2(y^2 - 5) = 0.$

For real solutions in $x$ , the discriminant must be non-negative:

$16 + 4y^2(y^2 - 5) \geqslant 0.$

Dividing by 4:

$4 + y^4 - 5y^2 \geqslant 0,$

$y^4 - 5y^2 + 4 \geqslant 0.$

Factorising:

$(y^2 - 1)(y^2 - 4) \geqslant 0.$

Since $y \geqslant 0$ , we can write this as

$(y - 1)(y + 1)(y - 2)(y + 2) \geqslant 0.$

As $y + 1 > 0$ and $y + 2 > 0$ for $y \geqslant 0$ , the sign is determined by $(y - 1)(y - 2)$ . This product is non-negative when $y \leqslant 1$ or $y \geqslant 2$ . Therefore either $0 \leqslant y \leqslant 1$ or $y \geqslant 2$ .

Part (ii)(b)

Very close to the origin: When $x$ and $y$ are small, the higher-order terms $x^4$ and $y^4$ are negligible compared to $x^2$ and $y^2$ . The equation

$y^4 - 5y^2 = x^4 - 4x^2$

becomes approximately

$-5y^2 \approx -4x^2,$

so $y^2 \approx \frac{4}{5}x^2$ , giving $y \approx \frac{2}{\sqrt{5}}x$ (taking the positive root since $y \geqslant 0$ ). The curve passes through the origin with slope $\frac{2}{\sqrt{5}}$ .

Very far from the origin: When $x$ and $y$ are large, the $x^4$ and $y^4$ terms dominate. The equation becomes approximately

$y^4 \approx x^4,$

so $y \approx x$ . The curve approaches the line $y = x$ asymptotically.

Part (ii)(c)

Differentiating $y^4 - 5y^2 = x^4 - 4x^2$ implicitly with respect to $x$ :

$(4y^3 - 10y)\frac{dy}{dx} = 4x^3 - 8x.$

Tangent parallel to the $x$ -axis ( $\frac{dy}{dx} = 0$ ):

$4x^3 - 8x = 0,$

$4x(x^2 - 2) = 0,$

so $x = 0$ or $x = \sqrt{2}$ (taking $x \geqslant 0$ ).

$x = 0$ : $y^4 - 5y^2 = 0$ gives $y^2(y^2 - 5) = 0$ , so $y = 0$ or $y = \sqrt{5}$ .
$x = \sqrt{2}$ : $y^4 - 5y^2 = 4 - 8 = -4$ , so $y^4 - 5y^2 + 4 = 0$ , giving $(y^2 - 1)(y^2 - 4) = 0$ , so $y = 1$ or $y = 2$ .

At $(0, 0)$ the tangent behaviour is degenerate (the curve passes through the origin with a definite slope $\frac{2}{\sqrt{5}}$ , but $\frac{dy}{dx}$ takes the indeterminate form $0/0$ ). Excluding $(0, 0)$ , the points with horizontal tangent are:

$(0, \sqrt{5}), \quad (\sqrt{2}, 1), \quad (\sqrt{2}, 2).$

Tangent parallel to the $y$ -axis ( $\frac{dx}{dy} = 0$ , equivalently $4y^3 - 10y = 0$ ):

$2y(2y^2 - 5) = 0,$

so $y = 0$ or $y = \sqrt{\frac{5}{2}}$ .

$y = 0$ : $x^4 - 4x^2 = 0$ gives $x^2(x^2 - 4) = 0$ , so $x = 0$ or $x = 2$ .
$y = \sqrt{\frac{5}{2}}$ : $y^2 = \frac{5}{2}$ , so $y^2(y^2 - 5) = \frac{5}{2} \cdot (-\frac{5}{2}) = -\frac{25}{4}$ . Then $x^4 - 4x^2 + \frac{25}{4} = 0$ , with discriminant $16 - 25 = -9 < 0$ , so no real $x$ exists.

Excluding $(0, 0)$ , the point with vertical tangent is $(2, 0)$ .

Sketch (first quadrant): The curve has two branches. The first branch starts at the origin (slope $\frac{2}{\sqrt{5}}$ ), passes through $(\sqrt{2}, 1)$ (horizontal tangent), reaches $(2, 0)$ (vertical tangent), then continues to $(0, \sqrt{5})$ (horizontal tangent). The second branch starts near the line $y = x$ for large values and comes down to $(\sqrt{2}, 2)$ (horizontal tangent), then connects to the asymptotic behaviour.

Part (iii)

The full curve for all $x, y$ is obtained by exploiting the symmetry of the equation. Since the equation involves only $x^2$ and $y^2$ (even powers), the curve is symmetric under $x \to -x$ and $y \to -y$ independently. That is, it is symmetric in both the $x$ -axis and the $y$ -axis (and hence also under $180^\circ$ rotation about the origin).

Reflecting the first-quadrant sketch into all four quadrants, the full Devil’s Curve consists of:

A closed loop surrounding the origin, passing through $(\pm 2, 0)$ and $(0, \pm\sqrt{5})$ , with four “indentations” at $(\pm\sqrt{2}, \pm 1)$ .
A separate component in each quadrant for $y \geqslant 2$ , approaching the lines $y = \pm x$ asymptotically. These four outer branches form a shape resembling a four-pointed star extending to infinity.

Examiner Notes

画图题需标注所有关键点(交点,切线方向变化点);x>=0,y>=0限制下曲线形状与全域不同;y的取值范围分析(0<=y<=1或y>=2)是解题关键。

Question 8

Topic: 纯数 | Difficulty: Challenging | Marks: 20

Problem

8 A pyramid has a horizontal rectangular base $ABCD$ and its vertex $V$ is vertically above the centre of the base. The acute angle between the face $AVB$ and the base is $\alpha$ , the acute angle between the face $BVC$ and the base is $\beta$ and the obtuse angle between the faces $AVB$ and $BVC$ is $\pi - \theta$ .

(i) The edges $AB$ and $BC$ are parallel to the unit vectors i and j, respectively, and the unit vector k is vertical. Find a unit vector that is perpendicular to the face $AVB$ .

Show that

$\cos \theta = \cos \alpha \cos \beta .$

(ii) The edge $BV$ makes an angle $\phi$ with the base. Show that

$\cot^2 \phi = \cot^2 \alpha + \cot^2 \beta.$

Show also that

$\cos^2 \phi = \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta} \geqslant \frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$

and deduce that $\phi < \theta$ .

Hint

(i) If W is centre of base, and M is midpoint of AB, then

$\vec{MV} = \lambda \begin{pmatrix} 0 \\ \cos \alpha \\ \sin \alpha \end{pmatrix}$

So a unit vector perpendicular to AVB is $\begin{pmatrix} 0 \\ -\sin \alpha \\ \cos \alpha \end{pmatrix}$

A1 (or B2)

Similarly, a unit vector perpendicular to BVC is $\begin{pmatrix} \sin \beta \\ 0 \\ \cos \beta \end{pmatrix}$

AS the obtuse angle between AVB and BVC is $\pi - \theta$ , the acute angle between the two unit vectors is $\theta$ , and so

$\begin{pmatrix} 0 \\ -\sin \alpha \\ \cos \alpha \end{pmatrix} \cdot \begin{pmatrix} \sin \beta \\ 0 \\ \cos \beta \end{pmatrix} = \cos \theta$

Hence $\cos \theta = \cos \alpha \cos \beta$ as required.

A1* [5]

(ii) $MW = VW \cot \alpha$ , $BM = NW = VW \cot \beta$ , $BW = VW \cot \varphi$ M1

By Pythagoras, $MW^2 + BM^2 = BW^2$ and so $\cot^2 \alpha + \cot^2 \beta = \cot^2 \varphi$ M1

$\tan^2 \varphi = \frac{1}{\cot^2 \alpha + \cot^2 \beta}$

and thus

$\sec^2 \varphi = \frac{1}{\cot^2 \alpha + \cot^2 \beta} + 1$

giving

$\cos^2 \varphi = \frac{\cot^2 \alpha + \cot^2 \beta}{\cot^2 \alpha + \cot^2 \beta + 1} = \frac{\tan^2 \alpha + \tan^2 \beta}{\tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta}$

$= \frac{\sec^2 \alpha + \sec^2 \beta - 2}{\sec^2 \alpha + \sec^2 \beta - 2 + (\sec^2 \alpha - 1)(\sec^2 \beta - 1)}$

$= \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \alpha \cos^2 \beta}{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \alpha \cos^2 \beta + (1 - \cos^2 \alpha)(1 - \cos^2 \beta)}$

$= \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \alpha \cos^2 \beta}$

$= \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$

M1 A1* [8]

$(\cos \alpha - \cos \beta)^2 \ge 0$

Thus

$\cos^2 \alpha + \cos^2 \beta \ge 2 \cos \alpha \cos \beta = 2 \cos \theta$

$\cos^2 \varphi = \frac{\cos^2 \alpha + \cos^2 \beta - 2 \cos^2 \theta}{1 - \cos^2 \theta} \ge \frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta}$

A1* [2]

$\frac{2 \cos \theta - 2 \cos^2 \theta}{1 - \cos^2 \theta} = \frac{2 \cos \theta (1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)}$

$1 - \cos \theta \neq 0$ as $\theta$ is acute.

Thus

$\cos^2 \varphi \ge \frac{2 \cos \theta}{(1 + \cos \theta)} = \frac{2}{(1 + \cos \theta)} \cos \theta$

As $\theta$ is acute, $(1 + \cos \theta) < 2$ and so $\frac{2}{(1 + \cos \theta)} \cos \theta > \cos \theta$

But also $\cos \theta > \cos \theta \cos \theta$

Hence, $\cos^2 \varphi \ge \cos^2 \theta$

As both $\cos \varphi$ and $\cos \theta$ are positive $\cos \varphi \ge \cos \theta$ , and so $\varphi \le \theta$ E1 E1 [5]

Model Solution

Part (i)

Set up coordinates with the centre of the base at the origin $W = (0, 0, 0)$ . Let the half-lengths of the rectangle be $a$ along i and $b$ along j, and let the height of the pyramid be $h$ . Then:

$A = (-a, -b, 0), \quad B = (a, -b, 0), \quad C = (a, b, 0), \quad V = (0, 0, h).$

Normal to face $AVB$ : The midpoint of $AB$ is $M = (0, -b, 0)$ . The direction from $M$ to $V$ is $(0, b, h)$ , which makes angle $\alpha$ with the base (the horizontal), so $\tan\alpha = \frac{h}{b}$ .

We need a vector perpendicular to face $AVB$ that lies in the plane perpendicular to $AB$ . Since $AB$ is along i, the normal has no i-component and lies in the $yz$ -plane.

Taking the direction from $M$ to $V$ as $(0, b, h)$ , a perpendicular direction in the $yz$ -plane is $(0, -h, b)$ . Normalising:

$\hat{\mathbf{n}}_1 = \frac{1}{\sqrt{b^2 + h^2}}(0, -h, b).$

This normal points inward (toward the interior of the pyramid). We can verify the angle: $\hat{\mathbf{n}}_1 \cdot \mathbf{k} = \frac{b}{\sqrt{b^2 + h^2}} = \cos\alpha$ , confirming $\alpha$ is the angle between the face and the base.

Normal to face $BVC$ : Similarly, the midpoint of $BC$ is $N = (a, 0, 0)$ , and the direction from $N$ to $V$ is $(-a, 0, h)$ , making angle $\beta$ with the base, so $\tan\beta = \frac{h}{a}$ .

A perpendicular to $BC$ (which is along j) in the $xz$ -plane is $(h, 0, a)$ . Normalising:

$\hat{\mathbf{n}}_2 = \frac{1}{\sqrt{a^2 + h^2}}(h, 0, a).$

Angle between the faces: The obtuse dihedral angle between faces $AVB$ and $BVC$ is $\pi - \theta$ , so the acute angle between the inward normals is $\theta$ . Therefore:

$\cos\theta = \hat{\mathbf{n}}_1 \cdot \hat{\mathbf{n}}_2 = \frac{1}{\sqrt{b^2 + h^2}} \cdot \frac{1}{\sqrt{a^2 + h^2}}(0 \cdot h + (-h)(0) + b \cdot a) = \frac{ab}{\sqrt{(a^2 + h^2)(b^2 + h^2)}}.$

Since $\cos\alpha = \frac{b}{\sqrt{b^2 + h^2}}$ and $\cos\beta = \frac{a}{\sqrt{a^2 + h^2}}$ , we have

$\cos\alpha \cos\beta = \frac{b}{\sqrt{b^2 + h^2}} \cdot \frac{a}{\sqrt{a^2 + h^2}} = \frac{ab}{\sqrt{(a^2 + h^2)(b^2 + h^2)}} = \cos\theta.$

Therefore $\cos\theta = \cos\alpha \cos\beta$ .

Part (ii)

Let $W$ be the centre of the base and $M$ , $N$ the midpoints of $AB$ , $BC$ respectively.

From the right triangle with vertex $V$ , centre $W$ , and midpoint $M$ :

$MW = VW \cot\alpha.$

Similarly from the right triangle involving $N$ :

$NW = VW \cot\beta.$

The edge $BV$ makes angle $\phi$ with the base, so from the right triangle $B W V$ :

$BW = VW \cot\phi.$

Now $B = (a, -b, 0)$ , $M = (0, -b, 0)$ , $N = (a, 0, 0)$ , $W = (0, 0, 0)$ .

In the right triangle $MBW$ (right-angled at $W$ since $MW$ is along j and $BW$ has components along both i and j… actually let me check):

$MW = b$ (along j), $NW = a$ (along i), and $BW = \sqrt{a^2 + b^2}$ .

By Pythagoras in triangle $MBW$ :

$BW^2 = MW^2 + NW^2 \quad \text{(since $MW \perp NW$)},$

$(VW \cot\phi)^2 = (VW \cot\alpha)^2 + (VW \cot\beta)^2.$

Dividing by $(VW)^2$ :

$\cot^2\phi = \cot^2\alpha + \cot^2\beta.$

Deriving $\cos^2\phi$ : From $\cot^2\phi = \cot^2\alpha + \cot^2\beta$ :

$\tan^2\phi = \frac{1}{\cot^2\alpha + \cot^2\beta},$

$\sec^2\phi = \frac{1}{\cot^2\alpha + \cot^2\beta} + 1 = \frac{\cot^2\alpha + \cot^2\beta + 1}{\cot^2\alpha + \cot^2\beta}.$

Therefore:

$\cos^2\phi = \frac{\cot^2\alpha + \cot^2\beta}{\cot^2\alpha + \cot^2\beta + 1} = \frac{\tan^2\alpha + \tan^2\beta}{\tan^2\alpha + \tan^2\beta + \tan^2\alpha \tan^2\beta}.$

Writing $\tan^2\alpha = \sec^2\alpha - 1$ and $\tan^2\beta = \sec^2\beta - 1$ :

Numerator $= \sec^2\alpha + \sec^2\beta - 2$ .

Denominator $= (\sec^2\alpha + \sec^2\beta - 2) + (\sec^2\alpha - 1)(\sec^2\beta - 1)$

$= \sec^2\alpha + \sec^2\beta - 2 + \sec^2\alpha\sec^2\beta - \sec^2\alpha - \sec^2\beta + 1$

$= \sec^2\alpha\sec^2\beta - 1$ .

Now multiplying top and bottom by $\cos^2\alpha\cos^2\beta$ :

Numerator $= \frac{\sec^2\alpha + \sec^2\beta - 2}{\sec^2\alpha\sec^2\beta} \cdot \frac{\cos^2\alpha\cos^2\beta}{\cos^2\alpha\cos^2\beta}$

$= \cos^2\beta + \cos^2\alpha - 2\cos^2\alpha\cos^2\beta$ .

Denominator $= \frac{\sec^2\alpha\sec^2\beta - 1}{\sec^2\alpha\sec^2\beta} = 1 - \cos^2\alpha\cos^2\beta$ .

Using $\cos\theta = \cos\alpha\cos\beta$ from part (i), so $\cos^2\alpha\cos^2\beta = \cos^2\theta$ :

$\cos^2\phi = \frac{\cos^2\alpha + \cos^2\beta - 2\cos^2\theta}{1 - \cos^2\theta}.$

The inequality: Since $(\cos\alpha - \cos\beta)^2 \geqslant 0$ :

$\cos^2\alpha + \cos^2\beta \geqslant 2\cos\alpha\cos\beta = 2\cos\theta.$

Therefore:

$\cos^2\phi = \frac{\cos^2\alpha + \cos^2\beta - 2\cos^2\theta}{1 - \cos^2\theta} \geqslant \frac{2\cos\theta - 2\cos^2\theta}{1 - \cos^2\theta}.$

Deducing $\phi < \theta$ : Factorising the right side:

$\frac{2\cos\theta - 2\cos^2\theta}{1 - \cos^2\theta} = \frac{2\cos\theta(1 - \cos\theta)}{(1 - \cos\theta)(1 + \cos\theta)} = \frac{2\cos\theta}{1 + \cos\theta},$

where we cancelled $(1 - \cos\theta) \neq 0$ since $\theta$ is acute.

Now we show $\frac{2\cos\theta}{1 + \cos\theta} > \cos^2\theta$ . Since $\theta$ is acute, $\cos\theta > 0$ , so this is equivalent to

$\frac{2}{1 + \cos\theta} > \cos\theta,$

$2 > \cos\theta(1 + \cos\theta) = \cos\theta + \cos^2\theta.$

$2 - \cos\theta - \cos^2\theta > 0,$

$(1 - \cos\theta)(2 + \cos\theta) > 0.$

Since $0 < \cos\theta < 1$ (as $\theta$ is acute), both factors are positive, so this holds.

Therefore $\cos^2\phi > \cos^2\theta$ . Since both $\phi$ and $\theta$ are acute (so $\cos\phi$ and $\cos\theta$ are positive), we conclude $\phi < \theta$ .

Examiner Notes

需正确找到各侧面的法向量;cos(theta)=cos(alpha)cos(beta)是经典立体几何结论;最后的phi<theta不等式证明需巧妙利用已得结果。